15.3 Double Integrals in Polar Coordinates
Claudia Castro-Castro
Math 283 Spring 2020
Double Integrals in Polar coordinates
- What if we want to interate over a region which description in terms of rectangular coordinates is rather complicated
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
- Some regions are easier to describe using other coordinate systems
Review of polar coordinates
- We have known that the polar coordinates \( (r,\theta) \) of a point are related to the rectangular coordinates \( (x, y) \) by the equations
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
- The relashionship between polar and rectangular coordinates is given by \[ \cos \theta = \frac{x}{r} \]
- \[ \sin \theta = \frac{y}{r} \]
- \[ \tan \theta = \frac{y}{x} \]
-
Polar to rectangular \( (r,\theta)\rightarrow (x,y) \)
\[ x=r\cos \theta \\y=r\sin \theta \] -
Rectangular to polar \( (x,y)\rightarrow (r,\theta) \)
\[ r^2=x^2+y^2 \\ \tan \theta = \frac{y}{x},\;\cos \theta = \frac{x}{r},\;\sin \theta = \frac{y}{r} \]
Polar plane
- The origin correspondes to a point in the plane called pole
- The polar axis is usually drawn horizontaly to the right and corresponds to the positive \( x \) axis
- Angles \( \theta \) are positve if measured in the counterclockwise direction from the polar axis and negative in the clockwise direction
- If \( r \) is negative, the point \( (-r,\theta) \) lies on the same line through the pole and \( (r,\theta) \)
Polar curves
- Graphs of function that relate to \( r \) and \( \theta \)
- Example: \( r=3 \)
-
Describes all points 3 units from the pole (or origin).
\( \theta \) is free - The curve \( r=3 \) describes a circle with center at the origin and radius 3.
Polar curves cont'd
- Example: \( \theta=\frac{\pi}{4} \)
-
Describes all points such that the line through them and the pole form an angle of \( \frac{\pi}{4} \).
\( r \) is free - If \( r<0 \), it lies in the quadrant on the opposite side of the pole
- The curve \( \theta=\frac{\pi}{4} \) describes a the straight line that makes an angle of \( \frac{\pi}{4} \) with the polar axis.
- More examples on the Wolfram Cloud Notebook
Polar regions
- Example: \[ D=\{(r,\theta)|\;1\leq r \leq 2,0 \leq \theta \leq 2\pi \} \]
Polar regions cont'd
- Example: \( D=\{(r,\theta)|\;0\leq r \leq 2,\frac{\pi}{4} \leq \theta \leq \frac{7\pi}{4} \} \)
Polar regions cont'd
- Example: \[ D=\{(r,\theta)|\;1\leq r \leq 2,0 \leq \theta \leq \frac{\pi}{3} \} \]
Polar Rectangles
- The region in the figure below is a polar rectangle
- \[ R=\{ (r,\theta)|\: a\leq r \leq b, \; \alpha \leq \theta \leq \beta\} \]
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
- How can we use polar representation to take advantage?
Double Integrals in Polar Coordinates
- Goal: compute the a double integral over a polar rectangle \[ \iint_R f(x,y)dA=\iint_R f(r, \theta)dA_{polar} \]
- Back to basis: Riemann sums
- Divide the interval \( [a, b] \) into \( m \) subintervals \( [r_{i–1}, r_i] \) with lengths \( \Delta r_i = r_i – r_{i –1} \)
- Divide the interval \( [\alpha, \beta] \) into \( n \) subintervals \( [\theta_{j–1}, \theta_j ] \) with lengths \( \Delta_j = \theta_j – \theta_{j–1} \).
- The polar rectangle is divided into smaller polar rectangles
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
-
\[ \iint_R f(r,\theta)dA_{polar}=\lim_{\Delta \theta \to 0 } \lim_{\Delta r \to 0 } \sum^m_{j=1} \sum_{i=1}^n f(r_i,\theta_j)\Delta A_{ij} \]
- What is \( \Delta A_{ij} \)?
- What is \( d A_{polar} \)?
Area of a polar rectangle
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
- Area of a circular wedge: \( A=\frac{1}{2} \Delta \theta R^2 \)
- \[ \Delta A_{ij}=\frac{1}{2}\Delta \theta \left[ \left(r_i+\Delta r\right)^2-\left(r_i\right)^2 \right] \]
- \[ \begin{align} \Delta A_{ij}&=\frac{1}{2}\Delta \theta\left(2r_i+\Delta r\right)\left(\Delta r \right)\\ &=r_i \Delta r \Delta \theta + \frac{1}{2}\Delta r^2 \Delta \theta\\ &\approx r_i \Delta r \Delta \theta \end{align} \]
- In the limit \( \Delta \theta \rightarrow 0 \) and \( \Delta r \rightarrow 0 \),
\[ \Delta A_{ij}\approx r_i \Delta r \Delta \theta \]
tends to
\[ dA=rdrd\theta \]
Example 1
- Find the volume of the solid bounded by the plane \( z=0 \) and the paraboloid \( z=1-x^2-y^2 \)
- The Cartesian plane intersects the paraboloid in the circle \( x^2 + y^2 = 1 \), so the solid lies under the paraboloid and above the circular disk \( D \) given by \( x^2+y^2\leq 1 \)
Example 2
- Find the volume of the solid that lies under the surface \( f(x,y) = \frac{1}{x^2+y^2+1} \), over the sector of the disk of radius 2 centered at the origin in the 1st quadrant.
- In rectangular coordinates the quarter of disk \( D \) is described by \[ D=\{(x,y)|\; x\geq 0, y\geq 0, x^2+y^2 \leq 1 \} \]
- In polar coordinates the quarter of disk \( D \) is described as a polar rectangle \[ D=\{(r,\theta)|\; 0 \leq r \leq 2, 0\leq \theta \leq \frac{\pi}{2} \} \]
Example 2 cont'd
- Transform the integrand by replacing \( x=r\cos \theta \) and \( y=r\sin \theta \) \[ \frac{1}{x^2+y^2+1}=\frac{1}{r^2+1} \]
- Replace \( dA \) with \( rdrd\theta \) \[ \iint_R \left(\frac{1}{r^2+1}\right)dA= \int_{0}^{ \frac{\pi}{2}} \int_0^{2}\left(\frac{1}{r^2+1}\right)\;r\;drd\theta \]
- Split into two integrals because we are integrating over a polar rectangle \[ \left( \int_0^{2}\left(\frac{r}{r^2+1}\right)\;dr \right) \left(\int_{0}^{ \frac{\pi}{2}} d\theta \right) \]
- Use substitution \( U=r^2+1 \) on the left integral \[ \left(\frac{1}{2}\ln{|r^2+1|} \right)\bigg\vert_{r=0}^{r=2} \left( \theta\right)\bigg\vert_{\theta=0}^{\theta=\frac{\pi}{2}} \]
- Evaluating we get the exact value of the integral \[ \frac{1}{2}\left( \ln{|5|-\ln{|1|}} \right)\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}\ln{5} \]
Other types of polar regions
- Just as in Cartesian coordinates, the domain of integration need not be part or a polar rectangle
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
- Combining the concept of type I/II regions from previous section, we can set up an integral in polar coordinates
- If \( f \) is continuous on a polar region of the form \[ D=\{ (r,\theta)|\; \alpha \leq \theta \leq \beta,\; h_1(\theta) \leq r\leq h_2(\theta)\} \]
- then \[ \iint_D f(x,y) dA= \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,r\sin\theta)\:r\:drd\theta \]
Example 3
- Find the volume of the solid that lies under the paraboloid \( z = x^2 + y^2 \), above the xy-plane, and inside the cylinder \( x^2 + y^2 = 2x \)
- The equation if the cylinder can be rewritten as \[ x^2-2x +y^2 = 0 \]
- Completing the square we get \[ x^2-2x+1+y^2=1 \]
- \[ \left(x-1\right)^2+y^2=1 \]
- Which in the Cartesian plane is the circle of radius 1 and center at \( (1,0) \)
- Tranforming to polar coordinates \[ \left(x-1\right)^2+y^2=1 \\ \Leftrightarrow \left(r\cos \theta-1\right)^2+\left(r\sin \theta\right)^2=1\\ \Leftrightarrow r^2\cos^2 \theta-2r\cos\theta+1+r^2\sin^2\theta=1\\ \Leftrightarrow r^2=2r\cos\theta\\ \Leftrightarrow r=2\cos\theta \]
- This circle is traced out on the interval \( -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} \)
- Thus the disk \( D \) is described by \[ D=\{ (r,\theta)|\;-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}, 0\leq r \leq 2\cos \theta\} \]
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
Example 3 cont'd
Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition
- We want the volume under the paraboloid and inside the cylinder \[ V=\iint_D \left(x^2+y^2\right)dA \]
- Transform the integrand by replacing \( x=r\cos \theta \) and \( y=r\sin \theta \) \[ \iint_D \left(x^2+y^2\right)dA=\iint_R \left(r^2\right)dA \]
- Replace \( dA \) with \( rdrd\theta \) \[ \iint_R \left(r^2\right)dA= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2\cos\theta}\left(r^2\right)\;r\;drd\theta \]
- Evaluate double integral \[ \begin{align} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2\cos\theta}r^3\;drd\theta&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{r^4}{4}\bigg\vert_0^{2\cos\theta}d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2^4\cos^4\theta}{4}d\theta \end{align} \]
- Use trig identity \( \cos^2 u = \frac{1+\cos2u}{2} \) \[ \begin{align} &=8\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1+\cos2\theta}{2}\right)^2 d\theta =2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[ 1+2\cos2\theta +\frac{1}{2}(1+\cos4\theta)\right] d\theta\\ &= 2\left(\frac{3}{2}\theta+\sin2\theta+\frac{1}{8}\sin4\theta \right)\bigg\vert_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}}=\frac{3\pi}{2} \end{align} \]